According to Maxwell speed distribution of molecules of gas at temperature $T$, most probable speed is given by

$$v=\sqrt{\frac{2kT}{m}} \, .$$

The corresponding energy is $$E= kT \, . $$

According to the Maxwell Boltzmann energy distribution, the most probable energy is $$E_p = \frac{kT}{2} \, .$$

Why are these different?

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    $\begingroup$ mean is not root mean square. Or let's say $E$ is not proportional $v$. $\endgroup$ Commented Apr 27, 2017 at 14:33
  • 1
    $\begingroup$ That would be because the mean of the square is not the same as the square of the mean. $\endgroup$
    – Pirx
    Commented Apr 27, 2017 at 15:22
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    $\begingroup$ and neither the mean nor the root mean square is equal to the mode, which is what the question actually asks about $\endgroup$ Commented Apr 27, 2017 at 15:26

1 Answer 1


Lets think about exactly what the Maxwell-Boltzmann distribution tells us. If we are being lazy we might say that it gives us the probability that a particle has a speed $v$, however since $v$ can take a continuous range of values, the probability that it takes any given value is $0$. Instead we say that it gives us the probability that a particle has a speed in a narrow range form $v$ to $v+\mathrm{d}v$. Now it is the probabilities, and not the probability densities, that physically matter and so should be independent of how we choose to represent them. Therefore $$ f(v)\mathrm{d}v = f(E)\mathrm{d}E $$ and not, as you might have naively though $f(v)=f(E)$. This is important because, since the relationship between $v$ and $E$ is non-linear, its sizes gets distorted as we make the transformation and $\mathrm{d}v\ne\mathrm{d}E$. This means that, even at corresponding energies and velocities, $f(v) \ne f(E)$. In fact, as @gautampk points out, since $v\propto E^{\frac{1}{2}}$, $\mathrm{d}v\propto \frac{1}{2}\mathrm{d}E$, leading to exactly the discrepancy you noticed.

This is related to a concept called the density of states, a function which counts the "number of ways" to have an energy in the range $E$ to $E+\mathrm{d}E$, or alternatively how much the length $\mathrm{d}v$ has been squashed and stretched. It is the $\sqrt{E}$ bit in the Boltzmann distribution for energy (along with some of the constants out the front). This is actually one of the rare moments where quantum mechanics makes life simpler. Our gas of particles is really quantum mechanical if we look closely enough and a gas of quantum particles in a finite volume has a discrete set of energy states. The density of states then really is counting how many of these discrete states have an energy in a certain range.

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    $\begingroup$ Specifically, the extra half comes about because $v\propto E^{\frac{1}{2}}$, so $\mathrm{d}v \propto \frac{1}{2} \mathrm{d}E$. $\endgroup$
    – gautampk
    Commented Apr 27, 2017 at 15:49
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    $\begingroup$ I would upvote this answer if gautampk's comment were incorporated into it. $\endgroup$
    – daniel
    Commented Apr 27, 2017 at 18:15
  • $\begingroup$ ...or let's say $E$ is not proportional $v$...cheers. $\endgroup$ Commented Apr 28, 2017 at 11:27

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